X J, Q (`$4`) K, A (`$5`) lose (`-$x`)
P(X) `8/52=2/13` `2/13` `9/13`

`E(X)=sum{x_i * P(x_i)}`

`=4times2/13+5times2/13-xtimes9/13`

`=frac{18-9x}{13}`

Now the expected value should be $0 for the game to be fair.

So `frac{18-9x}{13}=0` and this gives `x=2`.

So I would need to pay `$2` for it to be a fair game.

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